# Unproven claim in Weinberg's textbook

1. Sep 10, 2013

### dEdt

In his Lectures on Quantum Mechanics, Weinberg makes the following claim about solutions to the Schrodinger equation for a central potential: Suppose $\psi(\mathbf{x},t)$ is an eigenfunction of $H$, $\mathbf{L}^2$, and $L_z$. According to Weinberg, "since $\mathbf{L}^2$ acts only on angles, such a wavefunction must be proportional to a function only of angles, with a coefficient of proportionality $R$ that can depend only on $r$. That is, for all r,
$$\psi(\mathbf{x})=R(r)Y(\theta,\phi)."$$

He does not elaborate further on this, in my view, non-trivial statement. Can someone here provide a proof of his claim?

2. Sep 10, 2013

### Staff: Mentor

I'd be inclined to work on the intuition behind that claim instead of looking for a rigorous proof, because the general idea is useful in so many other problems.

But first, which part of the statement are you asking about? The part about $L^2$ acting only on $\theta$ and $\phi$, or the claim that this implies the $R(r)Y(\theta,\phi)$ form of the solution?

3. Sep 10, 2013

### dEdt

I'm talking about the claim that $\psi=R(r)Y(\theta,\phi)$ given that $\mathbf{L}^2$ acts only on angles.

4. Sep 10, 2013

### dextercioby

This is trivial once you consider the uniparticle Hilbert space for the 'dummy' particle 'moving' aroung the CoM. Its Hilbert space is L^2(R^3) = L^2 (R) X L^2 (R) X L^2 (R) and after performing a unitary transformation of the whole space you have that L^2(R^3) ~ L^((0,infinity),dr) X L^2 (S^2,dOmega).

In the position space the wavefunction of the 'dummy' particle would be then a product of a (sq integrable) function of r times a product of a function of spherical angles.